Q.Prove that $F(3) = 1 - e^{-9}$ if the probability density function $f(x)$ is defined as
$f(x) = \begin{cases} 3e^{-3x}, & x>0 \\ 0, & x \le 0 \end{cases}$

Cumulative distribution function (both cases). For any random variable $X$, the cdf $F(x)=P(X\le x)$ is defined for every real $x$.

• Discrete (Definition 11.4): $F(x)=\sum_{x_i\le x}f(x_i)$ — a step function, constant between support points and jumping by $f(x_i)$ at each $x_i$. Conversion both ways: given the pmf, $F$ is the running (cumulative) sum of $f$ up to $x$; given $F$, the pmf is recovered as the jump size $f(x_i)=F(x_i)-F(x_{i-1})$ at each point of discontinuity (with $F(x_0)=0$ before the first jump) — the jump of $F$ at $a$ is exactly $P(X=a)$.
• Continuous (Definition 11.7): $F(x)=\int_{-\infty}^x f(u)\,du$ — here $F$ is everywhere continuous (no jumps, since no single point carries probability). Conversion both ways: given the pdf, integrate piece by piece to build $F$; given $F$, differentiate — $f(x)=F'(x)$ wherever the derivative exists (at the finitely many "corner" points, $f$ may be set to any convenient value, since it never affects an interval probability). …